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For some numerical analysis of a fluid, I am wondering if there is any inequality that provides a lower bound (in the $L^2$ norm), for the $L^2$ inner product of a quantity with its derivative. In particular, I am seeking one that is useful for the form: $$(D(x),D(v))_{L^2} + (\nabla\cdot{x}, \nabla\cdot{v})_{L^2} + (\nabla\cdot{v}, \nabla\cdot{v})_{L^2}\ge C\|\nabla x\|^2_{L^2} + K\|\nabla v\|^2_{L^2}$$ for constants $C > 0$ and $K \ge 0$, where strain-rate tensor $D(v)\equiv \frac12(\nabla v + (\nabla v)^T)$, $t$ represents time, and velocity $v = \frac{d}{dt}x$ for displacement $x$.

I am trying to see if I can show that the left hand side is coercive in terms of $\nabla x$. i.e. it is okay if $K$ may be zero.

$x$ and $v$ are in the Sobolev space $V=\{\varphi \in (H^1(\Omega))^d \;|\; \varphi = 0\textrm{ on } \partial\Omega\}$ where $\Omega$ is a $d-$dimensional region of the fluid flow, for $d=2,3$.

P.S. If the inequality were to be reversed, I would have used the fact that $\|\nabla\cdot{v}\|_{L^2} \le \sqrt{d}\|\nabla{v}\|_{L^2}$ for a $d$-dimensional domain; and also applied Cauchy-Schwarz inequality, then Young's inequality, and Korn's inequality.

This answer on Math SE mentions Kantorovich inequalities, but there is no reference and my searches so far have been futile.

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    $\begingroup$ What are the domains of $x$ and $v$? $\endgroup$
    – user126920
    Commented Sep 17, 2018 at 1:06
  • $\begingroup$ Thank you, I have added more context to the question! $\endgroup$
    – Cogicero
    Commented Sep 17, 2018 at 1:23

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I don't think what you want is possible.

In phase space $x$ and $v$ are basically quantities that can be independently prescribed. The first terms of your left hand side is linear in $v$. So fixing $x$, $v$ arbitrary, for $\lambda$ of the correct sign and $|\lambda|$ sufficiently small, you will have that the left hand side is negative if you replace $v$ by $\lambda v$. A negative quantity can never be coercive!

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  • $\begingroup$ Thanks for your useful answer! I am inclined to accept this right away and move on, but a friend just suggested I could try integrating by parts on the first term on the left. Does this look like it makes sense? I'm clutching at straws here. Thanks! $\endgroup$
    – Cogicero
    Commented Sep 17, 2018 at 23:02
  • $\begingroup$ Also, I am sorry I see how the left is linear in $x$ so I think that infers linearity in $v$. However I can't see how the left hand side is negative. Could you please expand on that? Thanks! $\endgroup$
    – Cogicero
    Commented Sep 18, 2018 at 5:36
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    $\begingroup$ Essentially, you have some vector $A$ (which I use as a stand in for $\nabla x$) and some vector $B$ (which I use as a stand in for $\nabla v$). Your left hand side is basically $\langle A,B\rangle + \langle B, B\rangle$. Let $B' = \lambda B$. As long as $A$ and $B$ is not orthogonal, then you have $$ \langle A, B'\rangle + \langle B', B'\rangle = \lambda (\langle A, B\rangle + \lambda \langle B, B\rangle) $$ so for any $\lambda$ between $0$ and $-\langle A,B\rangle / \langle B, B\rangle$ the quantity is negative. $\endgroup$
    – Willie Wong
    Commented Sep 18, 2018 at 13:59
  • $\begingroup$ Thanks again. This is clear now. I have accepted the answer, but please I have one more related question. Does it change anything if the $\langle B,B \rangle$ term is not on the left i.e. the left hand is simply $(D(x),D(v))_{L^2} + (∇⋅x,∇⋅v)_{L^2}$ (that's what I get when I approach the problem a different way)? It seems to me, that it would mean $\langle A,B \rangle$ is on the left, which isn't necessarily non-negative. $\endgroup$
    – Cogicero
    Commented Sep 18, 2018 at 16:29
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    $\begingroup$ Without the $\langle B, B\rangle$ term the situation is worse. You don't even need $\lambda$ sufficiently small. If it has the "correct" sign it breaks the coercivity. $\endgroup$
    – Willie Wong
    Commented Sep 18, 2018 at 16:55

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